Question

A uniform chain of mass m and length / is lying on
a horizontal table with one third of its length hanging
over the edge of the table. If the chain is in limiting
equilibrium what is the coefficient of friction for the
contact between the table and chain?
​

Answer 1

Answer:Answer:

Given that the coefficient of friction is 0.5, assume that the length of chain on table be a and therefore length of chain which is left hanging be L – a.

Now to keep them equilibrium the force of friction be equal to force acted due to gravity upon the chain which will make it slide

So,  

                  F_f = F_g\\ => \mu N = mg\\ => \frac{1}{2} m^{'} g = m g\\ =>  as\quad the\quad linear\quad mass\quad density\quad equals\quad to\quad L / m\\ => \frac{1}{2} a \lambda = ( L -a) \times \lambda => a + \frac{1}{2} a = L\\=> a = 2L / 3